Red and Black(bfs)

Problem:Red and Black

Description:
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input:
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)
The end of the input is indicated by a line consisting of two zeros.

Output:
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input:

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output:

45
59
6
13

Language:C++

#include 
#include 

using namespace std;

struct node
{
  int x;
  int y;
}head,tail;

int W,H;
char s[21][21];
int step[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

void bfs()
{
  queue<node>q;
  int ans=1;

  q.push(head);
  while(!q.empty())
  {
    head=q.front();
    q.pop();

    for(int i=0;i<4;i++)
    {
      tail.x=head.x+step[i][0];
      tail.y=head.y+step[i][1];
      if(tail.x>=1&&tail.x<=H&&tail.y>=1&&tail.y<=W&&s[tail.x][tail.y]=='.')
      {
        s[tail.x][tail.y]='#';
        ans++;
        q.push(tail);
      }
    }
  }
  cout<<ans<<endl;
}

int main()
{
  while(cin>>W>>H&&W&&H)
  {
    for(int i=1;i<=H;i++)
    {
      for(int j=1;j<=W;j++)
      {
        cin>>s[i][j];
        if(s[i][j]=='@')
        {
          head.x=i;
          head.y=j;
        }
      }
    }

    bfs();
  }
  return 0;
}